算法刷题日志——dp

最长递增子序列

dp[i] 的值代表 nums 以 nums[i] 结尾的最长子序列长度。

class Solution {
    public int lengthOfLIS(int[] nums) {
        if(nums.length == 0) return 0;
        int[] dp = new int[nums.length];
        int res = 0;
        //dp[i] 所有元素置 1，含义是每个元素都至少可以单独成为子序列，此时长度都为 1
        Arrays.fill(dp, 1);
        for(int i = 0; i < nums.length; i++) {
            for(int j = 0; j < i; j++) {
                if(nums[j] < nums[i]) dp[i] = Math.max(dp[i], dp[j] + 1);
            }
            res = Math.max(res, dp[i]);
        }
        return res;
    }
}

最长公共子序列

class Solution {
    public int longestCommonSubsequence(String s1, String s2) {
        int n = s1.length(), m = s2.length();
        s1 = " " + s1; s2 = " " + s2;
        char[] cs1 = s1.toCharArray(), cs2 = s2.toCharArray();
        int[][] f = new int[n + 1][m + 1]; 
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                if (cs1[i] == cs2[j]) f[i][j] = f[i -1][j - 1] + 1;    
                else f[i][j] = Math.max(f[i - 1][j], f[i][j - 1]);
            }
        }
        return f[n][m];
    }
}

不相交的线

这是上一题的LCS的一个小变形

class Solution {
    public int maxUncrossedLines(int[] s1, int[] s2) {
        int n = s1.length, m = s2.length;
        int[][] f = new int[n + 1][m + 1];
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                f[i][j] = Math.max(f[i - 1][j], f[i][j - 1]);
                if (s1[i - 1] == s2[j - 1]) {
                //s1【i】,s2【j】都选的情况中就包含了f[i][j-1] f[i-1][j]
                    f[i][j] = Math.max(f[i][j], f[i - 1][j - 1] + 1);
                }
            }
        }
        return f[n][m];
    }
}